A light spring stretches 0.13 m when a 0.35-kg mass is hung from it.The mass is pulled down from this equilibrium position an additional 0.12 m and is released.Determine the spring constant k and the amplitude of vibration.
A)26.4 N/m and 0.37 m
B)26.4 N/m and 0.12 m
C)2.70 N/m and 0.25 m
D)26.4 N/m and 0.25 m
E)26.4 N/m and 0.13 m

Question

Quizplus · Accepted Answer

The spring constant k can be found using Hooke's Law: F = kx, where F = mg. Solving for k gives k = mg/x = (0.35 kg)(9.8 m/s²) / 0.13 m = 26.4 N/m. The amplitude of vibration is the additional displacement from equilibrium, which is 0.12 m.

A Light Spring Stretches 0